Theorem - Let $f:X \longrightarrow Y$ be a surjective map between the topological spaces $X$ and $Y$ . Then $f$ is an open mapping if and only if given a net $\{y_i\}_{i \in I} \subset Y$ such that $y_i \longrightarrow y$ , then for every $x \in f^{-1}(\{y\})$ there exists a subnet $\{y_{i_j}\}_{j \in J}$ that lifts to a net $\{x_{i_j}\}_{j \in J} \subset X$ such that $x_{i_j} \longrightarrow x$ . By "lift" we mean that $\{x_{i_j}\}_{j \in J}$ is such that $f(x_{i_j}) = y_{i_j}$ .

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Proof: $(\Longrightarrow)$ Suppose $f:X \longrightarrow Y$ is a surjective open mapping and $\{y_i\}_{i \in I} \subset Y$ is a net such that $y_i \longrightarrow y$ . Let $x$ be any element of $f^{-1}(\{y\})$ and denote by $\mathcal{N}(x)$ the set of neighborhoods around $x$ .

We define the set $J:= \{(U, i) \in \mathcal{N}(x) \times I : y_i \in f(U) \}$ and introduce in it the partial order given by $(U,i_1) < (V,i_2)$ if $V \subseteq U$ and $i_1 \leq i_2$ . With this partial order we claim that $J$ is a directed set.

Let $(U,i_1)$ and $(V,i_2)$ be two elements of $J$ . Since $f$ is open, $f(U \cap V)$ is a neighborhood of $y$ . Thus, since $y_i \longrightarrow y$ , there must be an $i_3 \in I$ such that $i_1, i_2 \leq i_3$ and $y_{i_3} \in f(U \cap V)$ . Hence, the element $(U \cap V , i_3) \in J$ and satisfies $(U,i_1) < (U \cap V , i_3)$ and $(V,i_2) < (U \cap V , i_3)$ , which proves that $J$ is a directed set.

A simple argument shows that the map $J \longrightarrow I$ given by $(U,i) \longmapsto i$ is increasing and cofinal, hence proving that $\{y_{(U,i)}\}_{(U, i) \in J}$ is a subnet of $\{y_i\}_{i \in I}$ .

Now, for each $(U, i) \in J$ pick an element $x_{(U,i)} \in U$ such that $f(x_{(U,i)}) = y_i$ (which exists, by construction).

It is clear that $x_{(U,i)} \longrightarrow x$ .

$(\Longleftarrow)$ Suppose now that the condition about nets stated in the theorem is satisfied. We shall prove that $f$ is then an open mapping.

Suppose $f$ is not an open mapping, i.e. there exists an open set $U \subset X$ such that $f(U)$ is not open in $Y$ . Thus, there is a point $y \in f(U)$ that does not lie in the interior of $f(U)$ . This implies that there is a net $\{y_i\}_{i \in I}$ such that $y_i \longrightarrow y$ and $y_i \notin f(U)$ .

Pick an element $x \in f^{-1}(\{y\})$ . By assumption, there is a subnet $\{y_{i_j}\}_{j \in J}$ such that $x_{i_j} \longrightarrow x$ , where $\{x_{i_j}\}_{j \in J}$ is such that $f(x_{i_j}) = y_{i_j}$ . But then, there must be a $j \in J$ such that $x_{i_j} \in U$ . This implies that $y_{i_j} \in f(U)$ , which is a contradiction. Thus, $f$ must be open. $\square$